The number which exceeds its positive square | vedclass

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Question

(B) Let the required number be $x$.According to the problem,the number exceeds its positive square root by $12$,so we have the equation:$x = \sqrt{x}

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) Let the required number be $x$.According to the problem,the number exceeds its positive square root by $12$,so we have the equation:$x = \sqrt{x} + 12$Rearranging the terms,we get:$x - 12 = \sqrt{x}$Squaring both sides:$(x - 12)^2 = x$$x^2 - 24x + 144 = x$$x^2 - 25x + 144 = 0$Factoring the quadratic equation:$(x - 16)(x - 9) = 0$This gives $x = 16$ or $x = 9$.Checking the values:If $x = 16$,$\sqrt{16} = 4$. $16 - 4 = 12$. (This satisfies the condition).If $x = 9$,$\sqrt{9} = 3$. $9 - 3 = 6 
eq 12$. (This does not satisfy the condition).Therefore,the correct number is $16$.

The number which exceeds its positive square root by $12$ is

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